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How does a glider in steady descent glide with negative AoA?
EDIT: I now understand relative wind is parallel to the travel of the glider. Ignorance was the basis of my earlier confusion.
I looked at some toy and commercial gliders and all of their wings were mounted with a 0 or a positive angle of incidence. This means that when a glider is gliding down, say at a glide angle of 5 degrees, wouldn't the angle of attack be -5 degrees? And wouldn't the negative AoA make the glider fall from the sky rapidly because of all of the down force? Or does the relative wind change?
I am making a small glider using the airfoil A18. I am confused because the glider doesn't seem to have any data below an AoA of 2. I am further confused because an angle of attack of 2 degrees has the highest Cl/Cd ratio which I thought usually happened at an AoA of 6-7. Furthermore, the Reynolds number is the highest which means that the wing creates turbulence which again I thought happened at a high AoA.
My glider will have a computer with an AHRS system that can control the elevator with a PID system so it can aim at one specific pitch, and I believe that pitch would be the glide angle. But to find the glide angle, I would need to know the lift and drag force which comes from the lift and drag coefficients. I don't have a wind tunnel so I can't calculate the lift coefficient at a different AoA. Plus, I'm launching at a high altitude (80,000 feet) so the lift force will change throughout the flight. Please tell me if I'm missing anything!
I also desperately need help on reading the graphs on AirfoilTools.com. Which graph is most important? Which polar should I read?
Why do these two different polars have different Reynolds number, with the same angle of attack? Is the Reynolds number something you input instead?
Am I overthinking this? Should there just be one angle, based on guess-timating, that my flight computer pitches down at, like -10 degrees? Or how would I gaurentee a stable AoA thoughtout my whole flight (probably 2 hours)?
Again, I am trying to make a sub-249 gram glider that is dropped, not thrown, from a high altitude. I welcome any tips.
Please excuse my ignorance and sorry for so many questions!

Roughly what chord and what airspeed?  That gives the Reynolds number.

If manueverability isn't crucial, why not just make it very statically stable like the "free flight" (uncontrolled) gliders common in 1930-1960, instead of fussing with PID?

Regarding Reynolds number, this might help

Re "I looked at some toy and commercial gliders and all of their wings were mounted with a 0 or a positive angle of incidence. This means that when a glider is gliding down, say at a glide angle of 5 degrees, wouldn't the angle of attack be -5 degrees?" -- why?

Way too many questions packed into one. You are aware that "free-flight" model airplanes, including gliders, have successfully flown for close to a century now, with no guidance of any kind, right? If not, google it.

How are you getting to 80,000'?

@quietflyer I'm using a weather balloon.

If you're really talking about weather balloons and AHRS, then you need a much better foundation in aerodynamics & aircraft design than we're going to be able to give you on a Q&A message board. It's okay to misunderstand what AoA is -- we've all been there -- but to leap from that point on the learning curve to creating new designs is pretty ambitious. May be best to slow down, study some basic aerodynamics, build & fly some gliders that are known to fly well, and build experience. Then you'll be in a better place to attempt the sort of project you're talking about here.

@RalphJ What happened was that I was doing all the electronics and the programming and the person responsible for the physical glider ghosted me, so now I need to do the glider, too. My friend is in charge of launching the balloon. I understand AHRS, Mahony filters, PID, etc. for glider flight.

@CharlesNicholson -- would be interesting to hear more.  If you accumulate 20 rep points you'll be able to talk in chat https://aviation.stackexchange.com/help/privileges/chat which you might find a better venue for general discussions.  Consider also the "Modeling Science" subforum of RCGroups -- https://www.rcgroups.com/modeling-science-136/ -- at least in the past some very well-informed contributors used to frequent that site.

In a glide, the relative wind is coming from below, giving a greater angle of attack than in level flight.
In a 5 degree glide with 0 angle of incidence and a nose-level attitude, the angle of attack is positive 5 degrees, not negative 5 degrees.
You don't want to pitch to the glide angle- that would give an angle of attack of 0. You want to pitch a few degrees above the glide angle.
Keep in mind that the lift and drag ratio for the airfoil is not the same as for the aircraft. A typical fuselage produces a lot of drag and very little lift. The best ratio for an airfoil tends to be at a smaller angle of attack than it is for a whole aircraft because the drag of the wings is a relatively small portion of total drag. So pitching up  produces a lot more total lift and not that much total drag.

Okay, so how do I make sure that the glider sustains the 2 degree angle of attack that the airfoil says it needs if the angle of attack can change? Does it do it by itself?

@CharlesNicholson It depends on what instrumentation you have. With GPS and vsi you could compute your glide angle and from there determine the required pitch angle. With an angle of attack sensor you could just use that. Or you could just rig your glider so that it's naturally trimmed around the desired angle of attack. Etc.

@CharlesNicholson Though also remember that the best angle of attack for the airfoil is probably not the best glide for the whole aircraft.

How would I rig my glider so it is naturally trimmed?

@CharlesNicholson That's a big question for a comment! It depends on the whole design of your glider. But a typical design has stability in angle of attack, and will tend to trend to one specific angle of attack. You can effect what angle of attack this is in many ways, such as changing the size or angle of incidence of the horizontal stabilizer.


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How to maximize L/D ratio by design in a high altitude glider?
EDIT: Some initial confusion with TAS and IAS confused me, I now understand this better, thanks for everyone who helped.
I'm designing a high-altitude glider, and I'm wondering how the angle of attack changes in a flight. I know a glider will go faster higher in the atmosphere. It's almost certain I'm lacking something crucial about aerodynamics.
Here are my questions:

Will a glider automatically adjust to the most optimum angle of attack for the given speed?
If the AoA changes in flight then the L/D ratio and hence the glide ratio must change, too, right?
The AoA must change in a high altitude flight, and how does one adjust for that?
How should one mount the wing to the fuselage if the AoA is changing in flight?

I'm horrible at communicating my questions because I don't really know why this isn't working for me, so I apologize in advance.

I've only flown power driven planes: when you say "adjust" for the change of angle of attack, do you mean something like trimming? The pilot usually makes changes in the pitch to get to the AoA needed to achieve whatever they want at whatever speed/altitude they're at.

Why are we voting to close a new user's Q, without a single word of explanation or comment about why it was insufficiently precise? Tsk, tsk, we shouldn't behave this was as a community. P.S. I don't disagree that the question could be tighter, but it would be kind to provide the OP some guidance before a smack-down.

The Perlan 2 is an excellent reference.  This glider has a crew of 2 and can reach an altitude of over 76,000 feet with a maximum glide ratio of 43.

@KennSebesta The system-generated explanation is pretty clear: Needs more focus This question currently includes multiple questions in one. It should focus on one problem only.

@RalphJ yes, and no. The "close" is quite abstract, and doesn't hint at what is too out of focus. Putting ourselves in the shoes of a brand new contributor, who doesn't have context gleaned from being here for years, what should the OP do to fix things? There are a literal infinite number of ways to modify the question, many of which will suffice, and many more of which won't. Without good feedback it verges on a "find me a rock" problem (https://jonathanbecher.com/2020/08/30/the-bring-me-a-rock-phenomenon/).

The boilerplate text also says "Update the question so it focuses on one problem only."  That's entirely clear. Right now, the question includes questions on "will the glider automatically adjust...", L/D and glide ratio changing, adjusting for changing AoA, and how to mount the wing to the fuselage. That's four questions. Focus on one. That's pretty clear direction from the system text.     As things stand, this question nearly asks for a multi-chapter textbook on design and flight, and that's wildly too broad to be answerable in this setting.

@RalphJ excellent feedback, thanks!

Will a glider automatically adjust to the most optimum angle of attack for the given speed?

Yes. Actually, it will adjust to the only possible angle of attack at a given speed. Within limits: If speed is too low, no angle of attack which creates enough lift will exist and the airplane stalls.
If the angle of attack is too high, more lift will be created exceeding weight and the airplane will begin to climb. Without adjusting of the pitch angle this will immediately lower the angle of attack until lift equals weight.

If the AoA changes in flight then the L/D ratio and hence the glide ratio must change, too, right?

Right.

The AoA must change in a high altitude flight, and how does one adjust for that?

Not necessarily. Normally, the adjustment for lower air density is done by increasing speed such that dynamic pressure will stay constant. At the same dynamic pressure, the same angle of attack will cause the same lift, so it can stay constant. What changes is the possible speed range.

How should one mount the wing to the fuselage if the AoA is changing in flight?

Pick the angle of attack for the wing which will be needed at the design point and then adjust the fuselage angle such that it will add the minimum amount of drag. Keep in mind that the angle at which the wing is now attached to the fuselage (called incidence) will also determine fuselage attitude during landing!
With a high aspect ratio wing you will see only limited changes in angle of attack, so the small variation of the fuselage angle will not dramatically change flow conditions around it. Pay more attention to the local pressure gradients to avoid early flow separation in the wing-fuselage intersection.

Let's keep it subsonic to help avoid significant changes in Lift/Drag ratio.
Take heart, this 1322 lb aircraft flew to 96,000 feet under power with an indicated airspeed of around 20 knots.  Using the True Air Speed calculator, you should have no problem with 50 knots IAS from 80,000 feet down to sea level.
Once you know the weight of the glider, design the wing area to lift that weight based on the lift equation:

Lift=density×area×coefficient of lift×True airspeed$^2$

As Peter suggested, build it with enough wing area to glide at 50 knots indicated airspeed without too high an angle of attack.
If you are really clever, make adjustments with wing size and weight until your best glide airspeed is around 50 knots.
You can make your glider staticly stable so it holds 50 knots in a glide.
Design iterations can be tested at the nearest hill.  If you are lifting your glider to altitude with a balloon, this is all very doable.  Include a tracking device (and maybe a camera too) in your glider and good luck.

Disclaimer:  please make sure you check with authorities before flying to 80,000 feet.

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How do I choose Cl during cruise flight with a given MTOM and variable S?
Currently I'm challenging myself to build an airplane from scratch. I have a background in (aeronautical) engineering but I have no experience in actually designing an aircraft from the ground up. (FYI: I want to build a flying wing - yes I know, I can already hear some of you asking why instead of a normal airplane but please just accept the fact that it's the most fun for me).
However, I am struggling to find a starting point. I figured it would be a good idea to start with the facts I have. Although not much, I know that:

The MTOM is going to be: $ m_{MTOM} \leq 250 \ \text{g} $ (for regulation reasons),
The cruise speed will (very roughly) be: $ v_{cruise} \approx [40 \dots 50] \ \frac{\text{km}}{\text{h}} = [11 \dots 14] \ \frac{\text{m}}{\text{s}}$ (I'm going to propel it with a either a T4x2x3 or T4x2.5 propeller powered by a 2S Li-Ion battery and a 1404 3800 KV BLDC motor),
The plane should be as efficient as possible with a feasible but scientific design process.

To get started on designing a suitable wing I thought it would make sense to select a fitting airfoil first. I read a lot about flying wings and what stuck with me is that for an (unswept) flying wing I need an airfoil with reflex. Three candidates I came across are:

PW75,
E230 and
E186.

Now I ask myself: how do I select at what $c_l$ I fly during cruise? How do I translate it to the 3D $c_L$ in order to calculate the surface area $S$ while in steady, horizontal flight ($F_W = F_L$)? What is the best planform for the resulting surface area $S$ to lift my plane? I.e. how do I choose $AR$ or $\lambda$ , $\varphi$ (sweep angle), etc.? Is this even the right starting point?
Edit 1: I settled myself on all types of flying wings, including swept flying wings.
Edit 2: I tried to make the question more constrained. So you don't have to list the entire aerodynamic design process of an airplane as an answer in order to answer it.

You mean something like a Northrop?

I plan on designing an unswept flying wing. More like the  Backstrom EPB-1. At RC scale it would probably come close to the Nano Goblin.

Update: yes, something like a Northrop as well!

This is an attempt to answer my own question.
From what I have learned so far, there are generally more ways than one to approach this problem. Generally speaking it will be an iterative process though.
To start working on my UAV I began to estimate $c_L$ in order to be able to calculate a planform $S$ that will lift the plane. For that we have to pick $c_l$, so the 2D lift coefficient, first. As @MaxPower suggested I started with a reasonable average $c_l$. In this case I start with $c_l = 0.3$. This results in $S = 0.0805 \ \text{m^2}$ at a cruise speed of $v_c = 13 \ \text{m/s}$ for the given $m_{MTOM} = 0.25 \ \text{kg}$. To get to $c_L$, again there is more ways than one, I used the equation in the top left corner of the table in this answer by @PeterKämpf. There is also a different approach by Lee Nicolai as @AeroAndy mentioned. I used the first one since it takes more variables into account, which is probably overkill in this application.
Since this approach uses more variables, you also have to have a clearer image of your plane in mind. In other words we need to know what aspect ratio ($AR$ or $\lambda$) to use and to what extend the wing will be swept (back) (sweep angle $\varphi$). As @MaxPower mentioned, it makes sense to use the highest feasible aspect ratio to minimize the induced drag (more specifically the induced drag coefficient). Consider @PeterKämpfs answer to see why $AR$ only influences the coefficient not the induced drag force directly. Since I will be 3D printing my plane (I am aware of skin friction and will handle this problem with post processing, etc.) I am constrained by the build volume of my printer and amount of segments the wing will consist of. I calculated different sets of wing spans and chord lengths for a rectangular wing planform and came to the conclusion that $b = 0.695 \ \text{m}$ and $c = 0.116 \ \text{m}$ at $S = 0.0805 \ \text{m^2}$, so $AR = 6$ seems feasible as a first iteration.
The sweep angle $\varphi$ is a bit more tricky however. Especially with flying wings since it's used for more than just having less drag at transonic speeds like in usual aircraft. In their book "Tailless aircraft in theory and practice" by K. Nickel and M. Wohlfahrt the authors give a lot of information regarding wing sweep. As, for many things, there is no ideal sweep angle. This is a bit outside the scope for this answer, however there is a trade off between efficiency of the control surfaces, directional stability, stall behavior, dutch roll tendency, etc.
This is a starting point to refine and iterate the planform.
Please leave a comment if there is something you disagree with or if I made a mistake, etc.
Edit: As a starting point I use a sweep angle of $\varphi = 15 \ \text{°}$. Now, I also try to find a more optimal planform by deviating from a rectangular wing to a trapezoidal wing. This is done by varying the taper ratio. This is also outside this scope. However, Nickel and Wohlfahrt come to the conclusion that the lower the taper ratio ($l_{tip} < l_{root}$) the worse the stall behavior. I start with a taper ratio of $\frac{l_{tip}}{l_{root}} = 0.8$. This is also closely linked to the sweep angle $\varphi$.

You might find Lee Nicolai's white paper "Estimating R/C Model Aerodynamics and Performance" a useful guide and reference. Look under "series resources/design event resources" at saeaerodesign.com

Thanks for the resource! I’m currently reading “Schwanzlose Flugzeuge:
Ihre Auslegung und ihre Eigenschaften”. It’s obviously in german but I can recommend it to anyone stumbling over this post. I’m learning a lot. Hope to some day be finally able to answer this question.

This is a great resource, however I'm finding myself already struggling at the basics. It feels to me kind of like the chicken or the egg problem to find the right planform area $S$, $c_l$ and appropriate wing loading $\zeta$. But to know how to get from $c_{l\alpha}$ to $c_{l\alpha}$ is huge. Also in the white paper the equation for $c_L$ is only valid for unswept wings. Do you know if it is still valid for small sweep angles e.g. up to 15 ° or even 20 °?

@eckh_ma, in their textbook, Fundamentals of Aicraft and Airship Design, Volume 1, Nicolai and Carichner assert a simple adjustment to the zero sweep lift curve slope fits empirical data well. For AR > 6, simply multiply the zero sweep slope by the cosine of the quarter chord (or maximum thickness line, if much different) sweep. For AR < 6 multiple by the square root of same.  This is in section 2.7, p. 44.

Thank you so much!

Also in my 2nd comment it should of course say:
"But to know how to get from $c_{l\alpha}$ to $c_{L\alpha}$ is huge."

I'd start calculating the lift curve created by the wing. It can be simply estimated from the 2D curve (the one of the airfoil) using for example the subsonic equation A2 that you find in the appendix of this NACA report:

In this equation a$_o$ is the slope of the 2D airfoil your wing is built upon; A is the aspect ratio and $\Lambda$ is the sweep angle; units of measurement are given at the beginning of the report. Use the correct Reynolds number when you choose the 2D airfoil plots. The maximum lift coefficient of your wing is also related to the one of the airfoil it is built upon and can be estimated with the following equation:
$C_{L_{max}}=0.9C_{l_{max}}cos\Lambda$
Obviously this equation doesn't consider any high-lift device.

What is the best planform for the resulting surface area S to lift my plane? I.e. how do I choose AR, λ, etc.?

You can play with the parameters in that equation to see how the lift changes accordingly. Anyway the biggest limitation of wing geometry is going to arise from the trim analysis (see later).

how do I select at what $C_l$ I fly during cruise?

The target lift is the one where your plane is going to fly the most. This really depends on how you are going to fly it. Once you have defined the lift $L$, you can get the relevant coefficient simply using its definition:  $C_L=\frac{L}{½\rho V² S}$. Obviously you have to check that your wing can deliver enough lift to takeoff, land and manoeuvre as well.

Is this even the right starting point?

Afterwards I'd perform a longitudinal trim and stability analysis: this answer should contain all you need for that; just set the contribution from the tailplane and the fuselage to zero (since you don't have them).

From the trim analysis you'll get that changing the pitching moment of the wing is the only way to trim your airplane: its value and its variation with the deflection of the control surfaces can be estimated with the methods contained in the USAF STABILITY AND CONTROL DATCOM (chapter 4.1.4 and 6.1.5 respectively);
From the longitudinal stability analysis you'll simply get that the CG must be in front of the quarter chord point.

As said, you can use the results from this last step together with the results from the lift's calculation to revise/design the planform of the wing: airplane design is an iterative process.

Chapter 4.1.5 of the DATCOM can be used to estimate the wing drag and therefore the needed power. I'd use a rear-mounted propeller since this location stabilises the aircraft in pitch and yaw.

Finally you can do a lateral trim and stability analysis to design the vertical tailplane.

Thank you very much for your answer! I understand that you want me to basically play with the subsonic equations in the second figure in the answer you quoted. Is capital lambda the same as the lower case one (taper angle)?
What are good ways to calculate a factor for Cl for take off and landing once I settled on a cruise Cl?
Do you know a good resource to learn about lateral trim and stability analysis?
I would also appreciate a source for why a rear prop stabilizes in pitch and yaw.
I'm new to this page so please point out if this is too much for a comment, etc.

I'd suggest some standard "general airplane design" textbook, like the one by Raymer or Roskam or Torenbeek. Note that those are all textbooks about full-scale airplanes: all methods/equations based on statistics are therefore strictly valid only for that class of airplanes (MTOW or CG estimation for example). Anyway what is based on physics/aerodynamics is still valid also for model airplanes (lift or drag or stability estimation for example). The DATCOM contains too methods to estimate the wing $C_L$ for straight subsonic wings, you can use it instead of the equation in the other answer.

Thanks! I will look into it.

Sorry, the capital Lambda is actually the aspect ratio. I did not convert the equations fully to that strange US habit of using different symbols when adopting equations.

Thank very much for clarifying @PeterKämpf

@PeterKämpf, I am kind of going in circles with this equation. These are some questions I have:

What exactly is this equation calculating? $c_{L \alpha 3D}$?
| 2. And is this supposed to be inserted into the equation for $c_L$?
| 3. Also is the capital L in the index not already saying it's the 3D lift coefficient (and vice versa for 2D)?

I would greatly appreciate some help because I am unfortunately not getting plausible data.

@eckh_ma I guess you mean the lift curve slope equation in the linked answer. You can simplify things by leaving out the Mach and sweep terms for your flying wing. The result is the derivative of the lift coefficient over the angle of attack in radians. Since it uses aspect ratio, it is for 3D lift. If you tell me your parameters and results, I may be able to help more.

@PeterKämpf, Exactly, sorry for not specifying better what I'm actually referring to. I fixed it now. It was a mixture of sleep deprivation and not converting degrees to radians... What I'm curious about now though, is why the difference between $c_L$ and $c_l$ is so small from what I calculated. And why $c_L$ is bigger than $c_l$ for every $\alpha$ (in other words $c_{L, \alpha} > c_{l, \alpha}$). Because I read quite often in the literature the the lift coefficient of a 3D wing is often smaller than its section lift coefficient. My parameters:  $AR=6$, $\varphi_m =0.262$, $Ma=0.04$, $\nu=0$.

@eckh_ma I have no idea how you calculate the 2D lift curve slope. It is about 2π plus some thickness correction. The 3D lift curve slope always has an aspect ratio related correction factor which is less than 1.

@PeterKämpf, Oh, I just used $c_{l, \alpha} = 2 \pi$ for the 2D lift curve slope. Is that not correct? So it's true that the lift curve slope for 3D, $c_{L, \alpha}$, is in fact larger than the 2D lift curve slope, $c_{l, \alpha}$, right?

@eckh_ma No, it always is smaller. Look at what aspect ratio does to the 2π in the 3D lift curve slope equation: It is always adding a factor of less than one.

@eckh_ma: chapter 4.1.3 (from page 494 of the pdf) of the USAF DATCOM contains everything you need.

@PeterKämpf, Okay that checks out with what I read. It was an error in my calculation (again)... It was very hard to find the (subtle) error. Thank you for your patience! However, I'm not quite sure what equation we are talking about now. Because in this answer in the equation for $c_{L, \alpha}$ in the top left of the table I'm not seeing $AR$ affecting $2 \pi$ (or $\pi$ for that matter) in the way you are describing. I think I'm misunderstanding...

@sophit, Thanks!

@sophit, As I'm reading your answer again, one question still remains: I agree with " [...] get the relevant coefficient simply using its definition: $c_L = \frac{L}{\frac{1}{2} \rho v^2 S}$." however, this still leaves me with two unknowns: $c_L$ AND $S$. I guess this highlights fairly well what I meant with "the chicken or the egg problem" when I commented AeroAndys answer. To know what $c_L$ I'm going to need, I need to know what surface area $S$ I require to lift the plane. But to know that, I need to know $c_L$ again and so on...

@eckh_ma You try to be the first airplane designer who does not need to iterate the solution? Please give up on that illusion! Pick one value, see where it gets you, improve the pick in the next iteration and so on. Regarding the top left equation: You do see AR in it, don't you? Next, enter the formula in a spreadsheet and calculate it for different ARs. What do you see?

@PeterKämpf, I'm sorry but I don't see where I said that I don't need to iterate. The question for me is how do I iterate and where do I start. Also, I struggle with judging the current iteration based on proven criteria because of missing experience and knowledge. That's the whole reason I turned to this website. I'm sorry if I made it seem like I was looking for some shortcut noone before me has ever found.

@PeterKämpf, Regarding the top left equation: I did evaluate it for different ARs. I see what it does. I'm just confused about how you see that "It is always adding a factor of less than one" at a glance. Because I'm not able to tell at first. I'm also still not sure about what you mean by "Look at what aspect ratio does to the $2 \pi$ in the 3D lift curve slope equation" because there is no $2 \pi$ in the top left equation. I hope you can help me resolve the confusion. Thank you for taking your time to help random people on the Internet learn about aeronautical engineering!

@eckh_ma: I expanded the section relative to the 3D lift, now it should be much clearer and easier to follow. The Northrop was based on a NACA 65-019, I'd start with that.

@eckh_ma Calculate the lift slope for an infinite aspect ratio. You will see that when you shorten the fraction properly, the equation is equal to π divided by ½. Which is equal to 2π in my book.

Thank you both so much @sophit and @PeterKämpf!

You state a plank wing but then go on to ask about planform and airfoil options, a plank is a flat rectangle.
So in general: Start with a reasonable average Cl for the class of airfoil you are using to get an approximate required wing area for basic structural design. If you use a pulling propeller the air flow will add airflow over the wing slightly increasing lift.
For subsonic aerodynamics an elliptical planform(more precisely elliptical lift distribution) is considered the most efficient, with the highest aspect ratio practical for the structural limits of materials, airfoil thickness ratio(bending and torsion strengths), construction methods, and mass distribution(eg Tanks/electric batteries inside wings vs fuselage). This minimizes induced drag. However, this is difficult to stabilize in a flying wing form factor, swept wings/delta are preferred for flying wings because the pitch control is easier (effectively blending a wing and tail together).
Use the Cl that occurs at the angle of attack which generates the best lift-to-drag ratio for your complete wing (The 2-dimensional airfoil optimum is minimum parasite drag, whole wing is induced drag plus parasite). Then scale the whole wing slightly up or down from the original area approximation to get the required total lift at the desired cruise speed. The best L/D can be obtained from a drag polar if you have accurate simulation or wind tunnel data that can be adjusted for Reynolds number.
Note this method does not necessarily account for other characteristics that may be desired such as takeoff and landing performance(including crosswind), maneuvering and gust safety factors(eg stall warning and spin recovery), oscillations, taxi-area wingspan limits, etc.

I appreciate your answer, thanks! I thought a "plank" referred to any unswept wing planform. I guess it also rules out taper. I'm going to rephrase the question.
I also thought of an elliptical planform or any planform with an elliptical circulation distribution. What do you think are reasons that the nano goblin doesn't have an elliptical planform besides ease of manufacturing?
Use the Cl that occurs at the angle of attack with the lift-to-drag ratio for your complete wing.
I would appreciate if you could elaborate this point since I don't fully understand it.

@eckh_ma That sentance was broken, I have edited it. …the Cl that occurs at the angle of attack which generates the best lift-to-drag ratio for your complete wing. So find the angle with the best lift to drag for your chosen wing, then calculate the cofficient of lift for that exact configuration. Finally use that Cl to scale the wing area to match your design operating weight. This is a somewhat iterative approach which may take a few cycles if there is a large scaling factor such as converting from small windtunnel models to human size aircraft.

Thank you very much! I'm still in the process of actually trying to do all the steps that you talked about (besides university and working) so that I can get a better understanding about everything.

So from what I understand you want me to: 1. Choose an average $c_l$ | 2. Convert $c_l$ to $c_L$ | 3. Calculate $S$ at desired cruise speed $v_c$ | 4. Redo the steps with the $c_l$ for best $\frac{L}{D}$. Is that correct?

@eckh_ma The order of steps is a matter of different perspectives. A builder starts at the foundation of a structure and finishes with the roof, because the bottom must support the top. An architect may start at the top with the required useful load because that determines the required strength and mass of the upper structure, the mass of the upper structure is required to calculate the load on the lower structure. To start engineering from the bottom would require many extra iterations. Opposite order for the same physical reason: the bottom must support the top. Airplane design is similar.

@eckh_ma regarding your old comment about the planforms and nano goblin. There is more to aerodynamics than simple efficiency. Behavior in adverse conditions is a major consideration. stalling at the wing root first is preferred to stalling near the tip or all at once. Ease of correct construction is a safety issue. The Reynolds number is very important, what is the scale of your design, 1 meter aircraft will behave very different from 10 meter at the same airspeed due to the boundary layer condition and relative boundary thickness changing the effective aerodynamic shape of the airfoil.

That's a good point! Can you please elaborate on "Ease of correct construction is a safety issue."? | I know that the Reynolds number is pretty important, however I only see it being factored it in when choosing the appropriate plot for $c_l$ vs. $\alpha$. From my point of view I still can't get behind the reasoning of $AR$ of the nano goblin. Is it really only that low because of maneuverability and behavior in adverse conditions? Do you think the final $AR$ is based on simulations or prototyping or both? What would you change about the planform of the nano goblin if anything?

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text:
How did the sailor survive getting sucked into an engine?

I'm sure we've all seen the famous video of the Navy sailor getting sucked into an engine.
Can someone explain how was this possible? I've read many comments but still don't have a visual idea how exactly was this possible. If you can use an engine diagram/image I would appreciate it for a better understanding.

No, we haven't all seen the video.  Could you summarise the event so that this question is complete in itself?  Questions should not depend on off-site resources to be understood.  In this case, you should identify the aircraft involved, and explain what happened.

The A-6 intakes are not that large. He did not go all the way down the pipe, but instead got wedged in. The sparks and whatnot coming out the exhaust was due to his headgear getting sucked off, and that went all the way in.

“What allowed him to survive was the design of the A-6 engine (the J-52). It has a long protruding ‘bullet’ or cone that extends in front of the first stage fans.
“When he was sucked in, his arm extended above his head which caused his body to wedge between the bullet and inside wall of the intake.”

https://theaviationgeekclub.com/video-shows-u-s-navy-crewman-getting-sucked-into-a-6-air-intake-he-survived-because-of-intruders-engine-design/

(image from the Smithsonian National Air and Space Museum)
https://airandspace.si.edu/collection-objects/grumman-6e-intruder/nasm_A19940152000

I'm interested in whether this was a design consideration, if that's answerable.

@JimmyJames - Given all the other fighter jets with much larger diameter intakes, I doubt "ground crew Ingestion prevention" was a design consideration. F-16/15/14....all have bigger pipe cross sections. There was at least one guy that was killed being sucked into an F-16. https://www.upi.com/Archives/1983/01/20/An-Air-Force-weapons-specialist-was-sucked-into-the/3085411886800/

Image taken from here
The reason that the petty officer survived is that the inlet duct is quite long (the whole body did not get sucked in, only from the waist up) and the engine stalled immediately because of the disruption of the airflow.

ohhh ok, so he never passed trough the engine blades????  I was totally unaware of that because i saw something accompanied with flames spitting from the back of the engine and I thought that was him.

@gabe the engine stalled for lack of air pretty quickly, but it still would have had fuel flowing briefly.  That fuel along with the sailor's loose hat will be in a hot environment at the tailpipe and igniting.  Just like an old automobike backfiring with excess unburnt fuel exiting via its exhaust.

@Gabe that was his helmet.

The issue is large pressure gradients around the engine inlet when the engine is operating at full power.  These create ingestion dangers around the intake as far away as 15-20 ft directly in front of the intake and extend to the sides of the intake as well.  If your body enters this space the gradient creates enough force on you to push you off your feet and draw you into the engine.
The video below (1:20 in the timeframe)is an example of a near tragedy which happened in 2006 launching a F-14D.  I’m not sure what happened here, but the shooter told the pilot to go to full power while the holdback petty officer was doing his final checks of the launch bar, Catapult shuttle, and hold back bar prior to launch. Note the way this guy ducks down to avoid getting sucked into the engine. I imagine the airboss chewed out that shooter for that!


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Are there any studies on the safety of glass versus steam instruments?
Has anyone come across any academic study on the effects of information provided by "glass" instrumentation that affects, favorably or unfavorably, situational awareness, skill and performance, and decision making?
Related question: Where does the term 'steam cockpit' come from?

On a related note, if you want to fly for a career, I suggest to everyone that asks me about it that they learn with steam gauges.  Chances are your first non-cfi job will be steam or EFIS and not full glass.  If you've never flown steam it'll be brutal.  Glass is easy, the big schools are doing their students a disservice by having all glass fleets.

@Ralgha Isn't EFIS considered glass??  (I agree with learning with steam gauges!)

It's a hybrid glass/steam arrangement.  Glass for the ADI and HSI, everything else steam (it can vary though).  Basically first generation glass, it's similar enough in layout to steam that you can pretty much switch between the two without thinking about it.

Before anyone asks a new question: http://aviation.stackexchange.com/questions/3577/where-does-the-term-steam-cockpit-come-from

The FAA released a 70 page report back in 2003 titled General Aviation Technically Advanced Aircraft FAA – Industry Safety Study which basically says that the overall accident rate of the two types of airplanes are almost the same.
They found that the "available safety" of Technically Advanced Aircraft (TAA) is greater than non-TAA aircraft, however in emergency situations pilots often were not aware of the resources available. To offset this, they recommend (among other things) more training for pilots before flying TAA's so that they actually have the safety related aspects available to them.
Here were their official findings:

Team Findings:

The safety problems found in the accidents studied by the Team are  typical of problems that occurred after previous introductions of new
  aircraft technology and all also reflect typical GA pilot judgment
  errors  found in analysis of non-TAA accidents
Previous safety problems similar to those identified in this Study have  been remedied through a combination of improved training and, in
  the  case of new aircraft capabilities, pilot screening (i.e.,
  additional insurance  company requirements of pilot experience).
The predominant TAA-system-specific finding is that the steps required to  call up information and program an approach in
  IFR-certified GPS  navigators are numerous, and during high workload
  situations they can  distract from the primary pilot duty of flying
  the aircraft. MFDs in the  accident aircraft did not appear to present
  a complexity problem. The  Team also believes that PFDs, while not
  installed in any of the accident  aircraft and just now becoming
  available in TAAs, similarly are not likely to  present a complexity
  problem.
TAAs provide increased “available safety”, i.e., a potential for increased  safety. However, to actually obtain this available safety,
  pilots must  receive additional training in the specific TAA systems
  in their aircraft that  will enable them to exploit the opportunities
  and operate within the  limitations inherent in their TAA systems.
The template for securing this increased safety exists from the  experiences with previous new technology introductions –the current
  aircraft model-specific training and insurance requirements applicable
  to  high-performance single and multi engine small airplanes. However,
  the  existing training infrastructure currently is not able to provide
  the needed  training in TAAs.
Effective and feasible interventions have been identified, mostly  recommending improvements in training, and effective implementation
  mechanisms for the recommended interventions exist. Therefore, TAA
  safety problems can be addressed, and the additional available safety
  of  TAAs to address traditional causes of GA accidents can be realized
  as  well.

APOA has a nice summary here.

Keep in mind that TAA != glass cockpit. A Technologically Advanced Aircraft must have: 1) an autopilot 2) an IFR GPS 3) an MFD that can display a map. The definition doesn't include anything related to glass instruments, unfortunately, which makes research hard.

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How much do aileron on a commercial airliner move for manoeuvring and trim?
I recently flew on an MD88 and had a good view of the aircraft's wing from my seat (read: I sat right next to the toilet). During the flight I noticed a couple odd things about the ailerons on the left wing. First, the middle one (there are three that I could see) was constantly pitched up by about an inch. Second, none of the ailerons seemed to move more than a couple millimeters during even the fastest banks (in fact, it was only that second aileron which I observed moving at all).
My questions are:

Was the second aileron being constantly pitched up for some reason? Or was it a trim issue?
How much do the ailerons even need to move to bank the aircraft? Would a couple millimeters do the job, or should I have seen more movement?

Edit: After taking some of the comments and searching around some more I found this diagram which seems to indicate what I was observing was not an aileron, but the outboard flap. Also this answer details why outboard ailerons are often locked in high-speed flight. But that still leaves the question of why the flap was constantly pitched up throughout the flight.

Surface control efficiency increases with dynamic pressure. Full deflection is only achievable at low speed (take off/landing) otherwise forces would be too important to fly safely (structural strength issues) and comfortably (roll rate)

That makes sense, but does it really only take a couple millimeters of movement once you're up to speed? Wouldn't that mean that the full inch of offset on the aileron should have had a dramatic effect on the plane's flight?

Are you sure you were looking at the ailerons and not the "multi-purpose" control surfaces? The ailerons are on the very end of the wing, the MP surfaces are on the top of the wing and closer to the wing root and towards the middle of the wing ahead of the flaps.

@thanby look for high speed aileron on this website to have a first high level overview.

@RonBeyer Distinguishing MP control surfaces from other control surfaces may be difficult if not knowing the aircraft type and/or the normal functioning of MP surfaces

Ailerons are the most outer controls, with a tab, I believe on the MD88, there is one per side (image and video). Other surfaces not on the trailing edge are spoilers/speedbrakes (raised at 5:10).

Considering the facts that the plane took off, maintained reasonably level cruise, and landed, all without incident, it seems to me that the couple of millimeters of deflection you saw were sufficient for the job at hand.

@FreeMan Yes, obviously things were working as intended (or being compensated for at the very least) because there was nothing wrong with the flight itself. I'm just curious about why I observed these behaviors.

Speedbrakes can also decrease (destroy) lift while adding drag. Decreasing lift on one side also allows to roll. Likely roll control is a mix of ailerons and speedbrakes.

@thanby The diagram you show is for a 727, not an MD-88. While probably pretty similar, I wouldn't use one to identify exactly what you were looking at on the other.

@RonBeyer I tried to find one of an MD88 but couldn't, unfortunately. The diagram looked pretty close (though the 88 doesn't have an inboard aileron like the 727) but you're right it might be different

@mins I think you're on to the right explanation but after looking at the definition of speedbrakes, I'm pretty sure that's not what I was observing. Speedbrakes seem to be akin to spoilers, while what I was observing was definitely a flap or similar.

Thanks for the diagram. That doesn't label the control surfaces unfortunately, but it could still help.

Here is a diagram of the DC9 (same than 88 except 2x3 flight spoilers/speedbrakes on the MD88 instead of 2x2) with designations.

@mins Ah perfect. Yes it was definitely the outboard flap that I saw deflected the whole flight.

I have never flown an aircraft, only flew in them, but I am studying aerospace engineering. If ailerons were deflected the whole time, I would say that it was a trim issue. Note that the MD-88 are old aircraft (introduced at 1980 and production stopped at 1999, source: Wiki page on MD80 series), so having some idiosyncrasies such as some roll instability or drift is to be expected. I have not studied the degradation of aircraft over time, but it makes sense. I had to mention that since stackexchange does not like people sharing opinions.
This question is more interesting for me since we can do some back of the envelope calculations to establish some simple facts. Lets approximate how much force would be required to get an angular acceleration of say 5 degrees per second squared, a very extreme case. Gradual turns are initiated with angular accelerations of about 2 degree per second squared; that way a 30 degree bank takes about 15 seconds to enter.
The most relevant thing that would be required is an estimate of the mass moment of inertia (MMOI) of the aircraft along its X axis (which is an axis pointing in the flight direction).
First the Maximum takeoff weight (MTOW) is taken from data (MD88 mass data) and the dimensions are approximated from drawings and data (MD88 dimension data)
Than the aircraft is modeled as a filled cylinder with the wings as two cuboid. The mass of these 3 sections are approximated from the mass fraction as found in this article. Crunching up all the numbers leads to an approximate $I_x$ for the MD-88 as $3350 \times 10^3$ $kg$ $m^2$.
We know that $M = I \alpha$, where $M$ is the rolling moment, $I$ is the MMOI and $\alpha$ is the angular acceleration. 5 degrees per seconds squared is about 0.09 radians per seconds squared. And $M = Fl$, where $F$ is the force from the aileron and $l$ is the moment arm. The moment arm is about 16 $m$ since the half wing span is about 17.9 $m$. So we have $F = \frac{I \alpha}{l}$. This leads to a requirement of 19000 N of force from both ailerons, and about 9500 N from each aileron.
In normal flight (at approximately MTOW) the wings are producing about 600 $kN$ of force to keep the aircraft in air. For this rather extreme banking maneuver the ailerons need to create a 10 $kN$ difference in lift (which is about 2% of the total lift).
Lets estimate the required change in lift coefficient for this change in lift. Assuming one aileron influences 15% of the wing surface, a flight at an altitude of 10 km, and a airspeed of 230 m/s, we find that each aileron should cause a 0.05 change in the lift coefficient of the wing segment. Have a look at the image below for an idea of about how much the lift coefficient changes with flap deflection. It seems evident that about a 5 degree deflection is required to get to the required order of magnitude of change in the lift coefficient.

(source: stanford.edu)
Now that we have some idea about the order of magnitudes, we can talk about the aileron deflection question. Using information from the aileron sizing documentation, we can think about the moment caused by your specific question. If we imagine a deflection of 5 mm (seems ok for stabilization but a little too low for control), we can approximate a aileron deflection of about 0.7 degree (assuming a $c_a/c$ of 0.25 and chord length of 1.5 $m$). Hence, I do not think deflections of a few mm are enough to roll the aircraft as would be required during landing maneuvers around an airport. Thus I think the aileron you saw that was trimmed is doing the heavy lifting for the rolling.
There are further subtleties introduced into the required magnitude of deflection of the ailerons due to other design parameters such as the incidence angle of the wing and the specific nature of the airfoil in the wing extremities. The former introduces a differential in how much the left and right ailerons need deflecting and the latter may cause the sections with the ailerons to have inherently higher lift in order to get a better lift distribution over the wing. These two are just to mention two of many things that interact to determine the rolling dynamics of the aircraft.
I wanted to include more sources for my work, but apparently I can only include two links; so I removed all of my supporting material and left only the image for consideration. If you want sources for my analysis, send me a message.

Now that is one heck of a good analysis and an excellent first answer. Welcome to the site!

If you want to put the sources in you can just put them in plain text. That way they're there and if somebody wants to they can go back and edit then into links

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Do flight recorders float in water?
In a recent BBC article regarding the disappearance of Malaysian Airlines Flight 370 (MH-370), the writer refers to a US Naval officer's statement:

Commander William Marks from the US Seventh Fleet, which is taking part in the search, says he expects the plane's flight recorders to be floating in the water.
"In calm seas, if there were a soccer ball [football] or a basketball floating in the water, the radar could pick it up. They [flight recorders] typically have a radio beacon and so for example our P3 [radar] - if they are flying within a certain range of that - will pick up that radio beacon. We have not yet picked up anything, but that's typically what those black boxes contain."

I was under the (potentially incorrect) impression that flight recorders, by nature of the materials needed to protect their contents in the event of a crash, are quite dense and unable to float in water. I base this impression on news reporting of other airliner crashes (such as Air France 447, which crashed in the Atlantic), where the recorders were found some time later on the ocean floor.
Additionally, it seems reasonable to me that one would want to have the recorders not float so that they would come to rest near the other debris on the ocean floor since the recorders are equipped with underwater locator beacons which would aid in locating not only the recorders themselves but also the debris of the plane. If the recorders floated then they could move significant distances from the crash site and other debris, complicating search efforts.
It seems that Commander Marks, the writer of the article, or I (or some combination thereof) misunderstand something about flight recorders and their behavior in water.
My questions are these:

Do flight recorders float on water? If so, why?
I know that flight recorders are equipped with underwater locator beacons. Are they also equipped with radio beacons (either an active transmitter or a some passive device like a corner reflector to make them easier to detect with radar)?

Edit: The article has been updated, with the section regarding floating data recorders now reading:

Commander William Marks from the US Seventh Fleet, which is taking part in the search, said he expected the plane's flight recorders to be floating in the water.
He said the recorders, also known as "black boxes", are fitted with radio beacons that can be picked up by radar.

they are painted bright orange with white stripes for easy visual detection

@ratchetfreak Yes, but do they float in water?

It would seem to me that they should float, it is trivial to have two gps/beaconing devices and put one on the floating black box and one on the plane somewhere. That way you can easily recover the black box and still have a beaconing sunken wreckage.

@TheZ Very little in aviation can be classed as trivial - especially anything requiring wholesale changes!

You are correct in the assumption that the typical data recorders being used today are too heavy to float.   They are designed to withstand very high forces, extreme temperatures, and to withstand corrosive environments (like sea water) for long periods of time.
For example:

A DFDR must obviously have a high degree of ‘crashworthiness.’ The
  first units were required to withstand a momentary shock force of
  1,000gs, the latest test standards now call for a test to 3,400gs for
  a duration of 6.5 milliseconds. The units also have to withstand a
  static crushing force at all of its six axis points of an applied load
  force of 5,000lbs for 5 minutes on each axis. Third, it must withstand
  a 500lbs piercing force test conducted by dropping it onto a ¼ inch
  steel pin from 10 feet. Lastly it must withstand a 1,100°C fire test
  for 60 minutes, and a 260°C oven test for 10 hours.
It is also required that these units are mounted within the tail area
  of an aircraft, away from the potential crushing force of any engines
  mounted nearby. The DFDR must be watertight to a depth of 20,000 feet
  in sea water, and survive at this depth for 30 days - and it must be
  fitted with an underwater locator beacon which will act like a sonar
  transmitter, by ‘pinging’ a signal through the medium of water that it
  might be laying in.

Time has an article showing the Air France 447 flight data recorders resting on the bottom of the ocean (there are others there):

This particular FDR is pretty common, and weighs about 13.5 pounds1.
Without seeing the original interview, I suspect that the author of the article got confused during the interview and was told that radar could detect debris the size of a soccer ball that is floating in the water and at a different time mentioned that they would be looking for the radio signals emitted by the black box (which wouldn't even be detected by radar).

UPDATE:
That being said, there are black boxes that can float and even eject from the airplane when a crash is detected.  They seem to be more common in military aircraft, and there have been a few attempts to require them in the US (and even have the government pay the $60,000/airplane price) but the bills stalled in congress.  The lawmakers here have recently started discussing this again though.
Deployable Black Boxes have been around since the 60's and are a second set of black boxes meant to be a backup to the main ones.  They can be particularly useful when crashing into water or in the event of a major fire because they would not be within the main crash site.

Here are a couple of articles about them:

Lawmaker Calls for 'Black Boxes' that Eject from Planes
Canada’s ‘deployable recorder’ would have located missing airliner within minutes, experts say

1 See specifications on Honeywell's webpage and page 7 of their product description.

Excellent. Thank you. I was trying to understand what reasons there might be for wanting a flight recorder to float -- it definitely seems like there was some misunderstanding with the Commander or reporter. Do you happen to have any information in regards to my second question (i.e. do flight recorders have radio beacons)? I've not found any information that says they do have radio beacons or radar retroreflectors, so I'm assuming they typically don't.

Well, they are normally inside the aircraft so a passive device wouldn't really help any.  As far as a radio beacon, aircraft are required to have ELT's which are radio beacons but the actual FDR's aren't.  They should be relatively close together though so that's probably why!

@ratchetfreak - A flotation device yes, one that could survive a 500kt impact and, possibly, ensuing fire...not terribly easy.  Granted, perhaps it could be worth it...

@Lnafziger Why not have 2 flight recorders, one floatable and another not? Also isn't 1000g enough? can a plane accelerate > 1000g?

@user13107 Actually, the standard used to be 1000g and they raised it to 3400g because it wasn't enough. Like I said in my first comment, I know that water landings have been considered but apparently the cost -vs- benefit ratio is not good enough for them to require it.  Even Air France 447's FDR was recovered, and that's pretty close to a "worst case scenario".

I suppose that the idea of the deployable black boxes is that you are able to auto-detect that something's wrong long before the impact, but long after the problems arise. Is this assumption correct?

@Lnafziger: Let AF pays for the search of the recorders, and they will feel this is more expensive than using appropriate devices in their 225 aircraft. MH370 recorders still not found, final cost still unknown.

The bills stalled... hehehe


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How does the runway number relates to Magnetic Heding/Course?
I know that runway numbers correspond to Magnetic North, how does it relate to Magnetic Course/heading?
Can we say that the runway number is equivalent to Magnetic heading?


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How Helicopters Change Attitude (Gyroscopic Precession or Thrust Vector Tilt)
I am trying to determine how helicopters pitch and roll (i.e. change attitude). I have heard two explanations.
1) Pitch/roll change is achieved by tilting the tip path plane and therefor tilting the thrust vector. By tilting the thrust vector, a moment is generated about the center of gravity which pitches/rolls the helicopter.
2) Pitch/roll is achieved by imparting a moment onto the rotor, which can be viewed as a giant gyroscope, causing the helicopter to precess with a 90 degree phase lag (see page 2-16 here).
In Helicopter Theory by Wayne Johnson (© 1980), section 5-5 pg 191, he writes:

Thus, there are two ways to view the rotor flap dynamics. First, the rotor blade can be considered a system with natural frequency of 1/rev, so that aerodynamic moments due to cyclic pitch excite the system at resonance. The response has a phase lag of exactly 90° (i.e. one-quarter of a cycle, which at 1/rev means an azimuth angle of 90° also). Alternatively, the rotor can be considered a gyro, with the flap hinges at the center of rotation forming the gyro gimbal. A control moment on the disk due to cyclic pitch then precesses the rotor with a 90° phase lag characteristic of a gyro until the flap damping produces a moment to stop the precession.

So, are the two explanations (1 and 2) equivalent, or are they both characteristics present? Wayne Johnson seems to suggest the former but gave no proof. Thanks in advance!

This animation shows it better than most.  The swash plate changes the plane of the rotor disc to change the plane of the lift vector.  It's the same thing as an airplane turning by banking.  When an airplane banks it imparts a lateral thrust component to the vertical lift component and it moves sideways, and because it's moving forward as it does so, it moves in a horizontal arc, or a turn.  A helicopter's rotor is just able to "bank" in all directions at any speed.
The swash plate directly flies the blades to the path that will put the overall disc in the desired plane of rotation.  The blades have to fly to their position using the air; if you had the machine running in a vacuum, the cyclic would have no effect.  However, gyro precession is a factor insofar as the input/result of the swashplate's effect on the blade's physical path is offset 90 degrees.
That is, to get the blade tip to its highest point in the circle, the highest blade angle, or highest lift force in the rotational cycle, is imparted 90 degrees before that point.  To tilt the rotor disc forward say, the blade that is on its way to its high spot over the tail boom will have reached its highest blade angle while passing at 90 deg on the left.
On the Bell rotor in the animation you can clearly see this.  The rotor disc angle follows the swash plate angle directly - swash plate tilts forward, rotor tilts forward. However, the linkage being driven by the swash plate acts on blade angle with a 90 degree offset (the control rods coming up on each side of the hub) to account for precession.
The Bell teetering system is interesting because it imparts no bending forces whatsoever to the rotor mast. The machine's body is suspended below the rotor hub as if it was a tennis ball on a string.  When you move the stick you are making the rotor disc fly around on its own account by tilting this way and that and the machine suspended under it follows along. In fact if a bending moment WAS applied to the mast, by having the teetering stops contact the mast while it's spinning, it comes apart and you become a falling object (this is "mast bumping" which cause the deaths of a number of Huey pilots in the 60s and Robinson pilots in the 90s).
Rigid and articulated rotor systems do impart some rolling and pitch moments into the air frame and the rotor head, mast and suspension system for the transmission are designed to transfer these loads.
A good fixed wing analogy between an articulated and teetering rotor is to compare regular airplanes vs hang gliders and trikes.  An articulated/semi rigid rotor is a bit like a regular aircraft where the body is directly acted on by the wings, although with a flexible connection between the wings and fuselage.  The Bell system is more like a hang glider or trike where the wing and body are on a fully flexible connection and you basically fly the wing around and the body underneath follows along without ever reaching the physical limit of the connection.  It's not a perfect analogy but sufficient for visualization.

your statement "The Bell teetering system is interesting because it imparts no bending forces whatsoever to the rotor mast" makes me think that both explanations account for how helicopters pitch/roll such that explanation 2 describes why the tip path plane tilts and explanation 1 explains how a roll/pitch moment is generated.

Also, I'm not sure if this is what you were suggesting, but a rotor in a vacuum would not precess as there would be no way to apply a torque to the rotor (via aerodynamic forces) to induce the precession.

Yes in a vacuum you'd have to apply the tilting force at the hub itself.

Explanation 2) is not an option to explanation 1). The page in the FAA handbook only points out that the rotor has 90° phase lag when applying forces for deflection. Has no consequences for the rotor cyclic pitch, tipping the rotor forward still speeds the helicopter up. It only has consequences for where to place the cyclic pitch force actuators.

the rotor is a large spinning mass, when you apply cyclic input, the lift simultaneously increases and decreases on opposite sides of the rotor due to change in blade pitch. When a moment is applied to a spinning mass, the mass will precess at an angle perpenducular (90°) to the applied torque and angular momentum vector. Precession is a behavior observed in gyroscopes. I would think they are connected in some way.

The 90º phase lag of a rotor has nothing to do with gyroscopes or precession, and is only due to resonance. A good explanation, by James Bennett, chief engineer of Cierva Autogiro Co. and a first-class expert in the field can be found here: https://archive.org/details/journal00britgoog/page/n136

please consider including the most relevant details of the paper here

@xxavier, thank you for the resource! I skimmed the first several pages and it seems to be similar to the explanation Wayne Johnson gives starting in section 5-5 pg 184, which concludes "By this means the pilot can control the attitude of hte helicopter, using cyclic pitch (swashplate tilt) to produce moments about the helicopter center of gravity by tilting the thrust vector". This is explanation 1 I offered above.

@Federico Please give the link...

@eball Direction control of the lift vector is always by tilting the disk, and that tilt is achieved (not always, but usually) via a swashplate or a spider. In the general case, for zero hinge offset, the control action on the rotor and the response lag by 90º. That lag is often explained as gyroscopic precession, also as 'aerodynamic' precession and as a resonance effect. The latter is (in my opinion) the best explanation...

@xxavier, so are explanations 1) and 2) equivalent (i.e. are they explaining the same physics)? I would question this conclusion because when you do not have zero hinge offset, the response lags by something less than 90º, which can only be explained by explanation 1) because gyroscopic precession always occurs 90º out of phase.

@eball It's easily explained by the 'resonance' hypothesis, because the existence of a non-zero hinge offset means that the blade is shorter, so has a different oscillation period. Hence, the phase lag is different from 90º...

@xxavier, I've never heard of the 'resonance' hypothesis and what you say may be true, but how does that explanation meld with explanation 2)?

@eball Pls see Helicopter performance, Stability and Control, by R.W. Prouty, an authority on the subject. Page 151...  https://b-ok.cc/book/5153040/71f1d8

@xxavier, thank you for the reference! Doesn't seem to address gyros but I only skimmed it. I'll have a deeper look when I have time. Thanks again for your help!


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How did Airbus A380 overcome corrosion issue in which aluminium wiring used?
The corrosion of aluminium is generally recognized to be a major problem. What new technologies or technics might be used in aluminium wiring to get away from the corrosion issue?.

No. I don't see any solution explained over there in regard to corrosion issue. Do you?

”Generally recognized"? It's an old wives' tale/urban myth/moral panic.  In other news, flying is a scary/dangerous way to travel.

@Harper Corrosion is a real problem if it's not properly solved.

"Do you think it might have been used [...]?" clearly asks for an opinion-based answer. Moreover, although the title is quite clear, the description is confusing and broad as it asks several questions.

What corrosion issue? What are you referring to?

@CrossRoads galvanic corrosion is the most common cause of corrosion for Aluminium

@ManuH I don't see anything wrong by asking and learning your opinion and scientific knowledge.

Yes. But I've not seen any stories reported on galvanic corrosion being an issue for Airbus. So what are you referring to?

You made me smile, cause I'm already asking what made Airbus have not an issue of galvanic corrosion. Corrosion issue is something that exists in stage of building an aircraft. If airbus flies, which's true,they did something not have the corrosion issue.

The A380 has not had any such wiring corrosion issues. Aluminium inside the insulation will not corrode because it is not exposed to ambient conditions. Copper corrodes too, but not inside insulation. Tons of planes are made of aluminium. The metal is treated to prevent corrosion. This is a non-issue, and it's perfectly safe, as answered in the question of which this is a dup.

My plane uses a big AL wire from the battery in the tail up to everything else in front of the pilot,and then copper everywhere else.  Big connector, crimped onto the wire and treated with something (I don't recall what anymore), it has lasted for 45+ years.

@huytergan as exlained in the help center, this is not a discussion forum but a Q&A website.

@ManuH okay, I'll fix it up

@CrossRoads and it's not even AA-8000 alloy (not invented yet) which is now a Code requirement in aluminum house wiring.

Your title and the body of your question ask two completely different questions. The one in the title seems eminently answerable; the one in the body is asking people to speculate about the future.

I have to agree with David. If you want to ask two questions, please do, as separate questions, not two in the same one.

The mechanisms by which aluminum corrodes are well-know and the measures needed to prevent it are similarly known to design engineers.
Aluminum corrosion via direct attack is an issue where the aluminum is exposed to air and the air contains moisture, and is made worse when salts are present in the environment and at certain pH ranges. It is prevented by encapsulating the aluminum in either an impervious coat of paint or plastic or a coating of intentionally-grown aluminum oxide. In an airplane, the insulation on the wire will prevent corrosion by direct attack.
Aluminum corrosion by galvanic action occurs when the aluminum is bonded directly to or in contact with a piece of metal that has a different position in the galvanic series, and the joint is wet or at least moist with water. You prevent it by excluding water from the joint by sealing the joint with glue or enclosing it in a housing that is airtight, and by coating the joint with a grease or oil that excludes water. In an airplane's wiring system, this would be done with a plastic plug-and-receptacle for the joint that keeps water out.
Electrical fires in wiring system connectors that contain aluminum wire joined with copper wire occur when the physical joint between them is vibrated or worked back and forth, which scrapes the native oxide off the aluminum in the joint, promoting more oxide growth which is scraped off, etc. which eventually leads to loading the joint with aluminum oxide and creating high contact resistance between the aluminum and the copper wire at the joint. I-squared-R heating at the degraded joint then sets the insulation on fire. This can also happen by temperature-cycling the joint which will cause the joint surfaces to be worked back and forth and once the joint begins to show resistance, then the heating occurs right inside the joint and the process accelerates itself.
In airplanes, houses, and cars where aluminum and copper wire must be joined, a special connector is used where the metals meet in which the crimp joint is exceptionally tight and the joint is loaded with a grease or chemical paste containing inhibitors that halt the oxidation of the aluminum.


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Chord definition misunderstanding
According to wikipedia's definition of chord (aeronautics), chord is a straight line that connects trailing edge to minimum radius point of leading edge.
What is so important about point of minimum radius that makes it the defining factor, or point to pass through, for chord definition? Why isn't chord, the line that defines longest distance from trailing edge to leading edge? That is the distance one giant caliper would measure on any wing?
It seems one slight modification to an airfoil can lead to massive chord angle variation with relation to that airfoil, while aerodynamic behaviour does(?) not look that much affected.

(two almost identical airfoils with minimal leading edge radius at various locations, resulting in large chord deviation. bottom is superposition of both)
If angle/line of zero lift of an airfoil looks useful for setting relative incidences between wing fuselage and horizontal stabilizer,
and the distance one giant caliper would measure looks useful -or seems the relevant value to pick- in order to define projected wing area,
I don't understand aeronautical chord's purpose and practical use.

It says the leading edge "may" be defined by the minimum radius of the leading edge.  How you choose to define it would make a difference in very curved turbine blade the article includes to illustrate this point, but I don't think it makes much difference in figures you use above.

"Minimum radius" is doubtful, for exactly the reason you demonstrate.  A more useful definition of LE is the forwardmost streamwise point, but that begs the definition of streamwise.  I haven't found anything authoritative though.

The definition of the "chord-line" is based on the leading and trailing edge as defined by the "mean camber line".

mean camber line - The locus of points halfway between the upper and lower surfaces [of the airfoil] measured perpendicular to the mean camber line itself. The most forward and rearward points of the mean camber line are the leading and trailing edges, respectively.
chord line - The straight line connecting the leading and trailing edges of the airfoil.
chord - the precise distance from the leading to trailing edge measured along the chord line.

From Introduction to Flight, John D. Anderson, Jr, McGraw-Hill, Copyright 1985, 1978. (Text from Aerospace Engineering curriculum, Texas A&M University)
To answer your question "What is the purpose of the chord line?", it is used to define the angle of attack seen by the airfoil. Which is defined as the angle between the relative wind and the chord line. In other words, it is an airfoil parameter (among others) used to describe an airfoil and characterize its performance.

You should consult Wikipedia with a healthy dose of skepticism.
Not that it isn't useful, but being the work of humans means that mistakes creep in from all sides. The chord definition is an example.
Normally, chord is the connection between the most forward part of the airfoil nose and the tip of the trailing edge, so your caliper method is as good as it gets on an uncambered airfoil. NACA Report 460, which published the original NACA 4-digit airfoil design method in 1933, only defines chord as

… the chord is taken along the x axis from 0 to 1 …

and what the x axis is can be seen in figure 1:

However, it gets more messy when camber is introduced: Now the chord line runs from the center of the circle inscribed into the nose radius to the trailing edge and is extrapolated until it connects with the airfoil contour:

This worked when airfoils were designed using simple equations, starting from a camber line and adding a thickness distribution. Since maybe 80 years, the design starts with a desired pressure distribution, with the contour only the result of that distribution, so there is no circle with a constant nose radius to start with. Pragmatically, we fall back to connecting the most forward and rearward part to find chord.

I don't understand aeronautical chord's purpose and practical use.

It is part of the wing geometry parameters. The precise angle and length is less important if its definition is used consistently. In the end, windtunnel or numerical simulation will give you the angle of the wing at which the desired lift is achieved, and its numerical value is of secondary importance.

To build up an airfoil, the classical way is to first draw the camber line (red) and then, along its length, circles with a radius corresponding to half the local thickness (grey). Then you connect the tops and bottoms of those circles with a smooth tangent (black).

And then there are airfoils like the GAW2, where the airfoil upper surface ends at z/c = -.0015, meaning the trailing edge end of the chord line is not on the airfoil and is just a random floating coordinate...

@ROIMaison In XFOIL, go into GDES and command NORM. A lot of coordinates are not cleaned up before put on a server.

Thanks for the pragmatic method, I see there are also 2 methods for drawing mean camber line. (both are called "camber line" but lead to different curves) First is the one described in David C. Rankin's answer and in your second graph. Second method is equidistant to upper and lower surfaces along Y axis, as seen here should one be preferred to the other?

@user721108 The canonical way is to draw the camber line and then to draw circles along its length with the appropriate radius to add the thickness distribution. Then you connect the tops and bottoms of those circles with a smooth line. This will only give you points perpendicular to the camber line when thickness does not change over chord.


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Am I correct about what AC/DC BAT BUS does?
The purpose of the AC/DC BAT BUS is to distribute either A/C electrical power and D/C electrical to the things in the aircraft that need them. For example, I think he whole ECAM system and displays use A/C electrical power. There are 2 of each BAT BUS (AC and DC). Do they work together so AC BAT BUS 1 and 2 work together and both distribute randomly or for example does AC BAT BUS 1 supply certain systems and AC BAT BUS 2 supplies the others. Do they have assignments of which systems to power, or do they work together?

On some aircraft the pilot/copilot displays and instruments are powered on completely separate busses, I can't say if the A320 is like this though.

See What does BUS mean in the startup procedure? which describes how buses are interconnected and can be switched on/off. This is not for Airbus but the principle is the same. (BTW you could improve your question by fixing typos and uppercase.)

You can't easily link the two AC buses together, you'd need to synchronize the phases first. And then a short on one side could cause a complete failure so you'd have to isolate them again. Way to complicated to get that failure tolerant. Instead they just use two independent AC systems. Generally each system exists twice and each one gets a different power source.
In the A320 you have two main AC buses (AC 1 and AC 2) in flight these are powered by engine 1 and 2 respectively. And there is a AC essentials bus, which is either connected to AC bus 1 or if that bus is unavailable it is connected to AC bus 2. The AC essential bus can also get power from the DC essential bus via essential transformer. So you can run the captain PFD and ND and the upper ECAM on battery power (only available above 80kts or sth. or in the air - for emergency only)
Check out the A320 "electrical system descriptions" and "indicating and recording" http://www.smartcockpit.com/plane/AIRBUS/A320.html
At the bottom of the indicating/recording systems you have a bus equipment list telling you exactly which consumer uses which bus. This is true for all other sections as well, so if you are interested in a particular system you can quickly find out where it is powered from.
According to A320 FCOM, 1.31.75 Indication/Recording Systems, Electrical Supply, Bus Equipment List

AC ESS bus powers the captain PFD and (via AC ESS SHED) the captain ND, the upper ECAM screen, the display management computer (DMC) 1 and
  3 (if selected and AC1 failed), flight warning computer (FWC) 1 and SDAC1. (and many many other things ...)
AC 2 powers F/O PFD, ND, lower ECAM, DMC 2, FWC 2 and SDAC2 (...)
AC 1 powers the DMC 3 normally (...)

On A320, the essential AC bus could be connected either to AC 1 or to AC 2, there is no AC BAT BUS 1 or 2.
The DC ESS BUS is connected to BAT 2.
HOT BUS 1 is connected to BAT 1, HOT BUS 2 is connected to BAT 2.

@fooot, thank you for the edition


text:
What does 'given rate' mean?
I was watching this video:

At around 1:28 I heard the controller clear a plane to descend to a level "given rate". I was searching online but couldn't seem to find what they meant with the term 'given rate'.
Could it be the "level change" or "open descend" button in the aircraft?

Sometimes, a controller will give you a specific rate of descent along with descent clearance. Example: "Descend FL140, rate 1500 feet per minute or more". When a new descent clearance is given, "given rate" indicates that the previous rate of descent restriction still applies. In this example, if a subsequence clearance was given: "Descend FL80, given rate" it would mean the same as "Descend FL80, rate 1500 feet per minute".

Appreciated! Got any source on this? (It makes sense but still) :)

I'm actually not sure whether it is really standard phraseology. You could always check out CAP 413.

The ARTCC Controller is telling a flight crew currently descending to descend to FL150 at their current rate of descent e.g. 2000 fpm, etc. either to expedite their current descent or the controller is satisfied with the current rate of descent to traffic separation and flow into the terminal area where the flight will be landing.

ARTCC is a FAA local terminology, Europe uses the ICAO wording ACC (area control center). Here probably Langen ACC (EDGG) in Langen FIR.

Got any source on this? Thanks!

If the controller wanted the "Current Rate"  think he would have said so.
I have flown into Frankfurt many times and have never heard the phrase "Given Rate" used. It may have been a short term phraseology trial, but it also makes sense to me that the pilot may have been told earlier to descend at specific rate of descent and that the controller was again asking to descend at the same "given rate".

It might not be perfect phraseology, but that’s what the controller meant.

'Given rate' is non-standard phraseology. The 'given rate' refers to a 'rate of descent' given in a previous instruction.

Hello and welcome to aviation.se. While your answer may be correct it is very short and of rather poor quality. It would be greatly appreciated if you could provide some background information and ressources to your post.


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What is the difference between a positioning flight and a ferry flight?
What's the difference (for the purpose) between positioning and ferry flight?

A positioning flight is a flight for the sole purpose of positioning the aircraft to conduct another flight from another airport. This is often done when the aircraft finishes its day in one city, but is needed in a different city the following day because another plane has broken down.
A positioning flight is technically a type of ferry flight, however the latter is a more broad term. Examples of ferry flights include delivery from the manufacturer, or flying to a different city for heavy maintenance.
Both flights are generally conducted without fare-paying passengers on board, but sometimes extra airline staff are carried.

A "ferry flight" can also have a regulatory connotation in FAA-land: It can refer to transporting an aircraft under a "ferry permit" - or more precisely a "special flight permit" - when it might not currently meet airworthiness requirements (one example is an aircraft with an expired annual being flown to an airport where the annual can be conducted).

Why don't they charge money for these seats? Surely they have to make the plane just as safe whether there's paying customers on board. Seems like they're leaving money on the table!

@corsiKa, "these seats" are their own (or contractor, e.g. maintenance) staff they need to transport. By putting them on a positioning flight they are saving money as they would otherwise take up space on a revenue flight.

@corsiKa Occasionally they do take 'overflow' passengers on positioning flights. But they are usually arranged on such short notice you couldn't sell anything if you tried, plus then you need to pay for cabin crew. And as voretaq7 excellently points out, some ferry flights are not (legally) as safe as normal ops so passengers are not allowed.

@corsiKa Just some examples of the very-short-notice type of flights Ben spoke of. Three of the many ferry flights I did were because of pressurization problems, which meant we were required to stay at or below 10,000 feet. Two were 3-engine 747 ferry flights, which meant that we were not allowed to have more than the required crew on board: pilot, co-pilot, and flight engineer (old 747-200s).

Ferry flight refers to flying the aircraft from the factory or flying of the aircraft to or from major maintenance (overhaul).
Positioning flight refers to the aircraft to some place(airport) from which it is operated. For example, an aircraft may be positioned at an airport for normal operations from the next day.
Both are non revenue flights. They are often used interchangeably. NTSB, however, seems to consider ferry flights as a type of positioning flight.

Eurocontrol defines positioning flight as:

A non-revenue flight carried out to position an aircraft for a scheduled or non-scheduled flight or service.

There's no absolute accepted definition across the world, however generally it depends on whether passengers could be carried on the flight. In a positioning flight the airplane could carry passengers, in a ferry flight usually not due to the condition of the airplane (mechanical problems, extra fuel tanks) or the ownership.

Positioning flights do not have to be approved, whereas ferry flights have to be approved by the FAA by permit or operations specifications.  Due to the nature of ferry flights only required crew and or mechanics my fly on the aircraft due to airworthiness limitations.

This answer could be much stronger if a few sources were added.


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What is the name for a flight crew that includes extra members for a longer flight?
For short haul flights, airlines fly with BASIC CREW.
For long haul flights, the crew must be composed for another crew or "extra" crew members, less than a crew.
What's the designation of the crew, that comprises more than the minimum number required to operate the aeroplane?
For example: 2 captains and 2 FOs; 1 relief pilot; 2 captains and 1 FO.

What do you mean the "designation"? Are you asking what the parameters are that dictate the number of crew members? Are you asking about flight deck crew, or cabin crew?

I meant, according to the number of crew members, there is an specific designation for them, as I have written, BASIC CREW for the minimum required number of crew members. In long haul flights, crew members can leave their assigned post, and be replaced by another appropriately qualified flight crew member for the purpose of in-flight rest. Some flights have 2 crew on board, including pilots and cabin crew members, and in others, one or two extra crew, for the same purpose. The designation for these teams, as BASIC CREW for the single ones.

The number of cockpit crew required is based upon length of flight.  Cabin crew numbers are based upon number of passengers as minimum but there might be extra to meet quality standards they wish to meet for first and business class customers.  The exact regulatory minimums will be based upon your jurisdiction, so please include in your question tags the regulatory body you want an answer for (e.g. faa, easa, etc).

ICAO is the agency.

ICAO is not regulatory, they only do recommendations

My country follows ICAO recommendations, that's why I asked about ICAO.

FWIW, back in the 1990s it was referred to colloquially as a "heavy crew."

A crew that consists of the basic required cockpit crew for the aircraft plus relief crew for long flights is called an "augmented flight crew." As far as I know, this designation is used by ICAO, the FAA, and the EASA (and likely others.)
This ICAO document defines augmented flight crew as:

Augmented flight crew. A flight crew that comprises more than the minimum number required to operate the aeroplane and in which each flight crew member can leave his or her assigned post and be replaced by another appropriately qualified flight crew member for the purpose of in-flight rest.

The FAA's Federal Aviation Regulations define it nearly identically in 14 CFR 117.3:

Augmented flightcrew means a flightcrew that has more than the minimum number of flightcrew members required by the airplane type certificate to operate the aircraft to allow a flightcrew member to be replaced by another qualified flightcrew member for in-flight rest.


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Are propellers obsolete?
Now that we have jet engines and all, is there still a use for propeller based flight (besides helicopters)?
Most airlines nowadays don't use propellers, right?

It depends on what you name a jet engine and what you name a propeller, On a high dilution turbofan, the fan is a propeller in addition of being the first stage of the compressor. Most of the thrust is obtained by the fan-accelerated air which doesn't travel through the engine core. Actual turbojets are not used in commercial aviation (cruising at Mach 0.8) today (Concorde used turbojets).

Propellers are not obsolete by any measure.
The jet engine offers two important advantages over propeller aircraft- higher speed and better fuel economy over long range. However, for low speeds and shorter routes, propeller (tuboprop) aircraft are better.
Also, for light GA aircraft, jet aircraft is prohibitively expensive to purchase and maintain. Among other things, the higher speeds associated with jet aircraft make them unsuitable for use in GA. Also, every jet pilot is first trained in propeller aircraft; in fact, airline pilots should have a minimum prior experience before piloting jet aircraft.


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Did A320 NEO need an equivalent to 737 MAX MCAS because of its bigger engines?
We read that MCAS on the Boeing 737 MAX was needed because of the new, bigger, heavier engines of the MAX compared to the previous 737 versions.
The closest equivalent to MCAS on the Airbus A320 NEO is the alpha protection even if they are not the same.
But what have been the adaptations to the A320 due to heavier, larger engines?  Can we be sure that A320 does not suffer from issues comparable to those the 737 MAX has, due to their similarities (bigger engines on same airplane)?

You're going to get shot down here due to having 6 questions in one request.

And your premise is not correct. It wasn't just "new, bigger, heavier engines", it is their change in location "The new Maneuvering Characteristics Augmentation System mitigates the pitch-up tendency of the new flight geometry due to the heavier engines being located further forward and higher than the previous ones."

"Can we be sure that A320 does not suffer from issues comparable to those the 737 MAX has, due to their similarities" looks awfully close to accident speculation, which is off topic here.

The A320 and 737 have very different flight control architectures.
The 737 has physical cables that transmit pilot (or autopilot) input directly to the hydraulic actuators. This was common in the 1960's when the airplane was first designed. This means that the airplane handling comes down to the aerodynamics and the pilot input.
The 737 MAX presented an issue when it was found that certain situations approaching stall could cause the airplane to pitch up, exacerbating the stall, rather than naturally pitching down and recovering as expected. This was due to the larger engine nacelles, which are located ahead of the center of gravity. An aerodynamic change would have taken lots of time and money and had its own drawbacks. Instead, Boeing created MCAS. This applies trim to help to pitch the airplane down when it reaches high angles of attack.
The A320, on the other hand, was designed as a fly-by-wire aircraft. This means that a computer is using sensor inputs to control the airplane all the time, and pilot or autopilot inputs only provide a guide to that computer control.
Fly-by-wire allows some helpful safety features, such as not allowing the airplane to exceed the design envelope, including stall. This means that any A320 in normal conditions will allow the airplane to reach a maximum angle of attack and no higher. Any aerodynamic differences such as a different fuselage length, larger engines, etc, are countered by the flight computer. Other than some possible fine tuning of the control laws of the fly-by-wire system, the A320 neo would not need any special new system to control angle of attack.

[1] Neither this, nor the answer posted by @Barath are not actual answers. There are several issues: (1) Is the 320 ceo inherently less aerodynamically stable than the 737 NG? (2) Do bigger engines introduce similar pitch-up issues on the Neo - aerodynamically, not in piloting - as they did on the Max? (3) Does neo mitigate these issues using existing fly-by-wire system without affecting piloting? (4) How many computers the neo has? (I believe the ceo has 3 independent 286-based computers. The Max has two computers). (5) Can the Neo switch between computers in-flight? (The Max cannot).

[2] (6) How many sensors does each Neo computer reads from? (The Max reads from a single AOA and a single airspeed sensor on one side of the aircraft; sensors on the other side are connected to the second computer, that is turned off in flight). (7) What is the design and positioning of the sensors on the Neo compared to the Max, are they better protected from impact or more durable? (8) When the fly-by-wire system activates on the 320, how much freedom does it give to a human pilot? (The MCAS is explicitly designed to overpower the pilot).

@RustyCore Welcome to Aviation.SE! I would suggest you ask new questions, at least splitting the architecture (how many computers, how many sensors, location of sensors) and the functionality (FBW effect on handling, pilot input) into separate questions. We also have some related questions already, such as this and this, feel free to look around and reference existing questions if there's something more you'd like to know.

The 737 Max had short landing gear. Since the engines need to have a minimum ground clearance, this meant that the new larger engines had to be repositioned further forward and higher on the wing.
The repositioned engine along with the new engine nacelle shape meant it had different flying characteristics at high angles of attack (e.g. due to airflow). Thus Boeing introduced the MCAS system to reduce chances of stall in manual mode and try to make the plane behave the same as before.
The A320 has larger "legs". It could accommodate the larger heavier engines of the A320neo without requiring re-positioning from the A320ceo. In fact, the LEAP 1A of the A320neo is marginally larger and heavier than the LEAP 1B of the 737Max. Even the PW1100G option for the A320neo is marginally larger than the LEAP1B (737MAX).  [All the newer gen engines are larger and heavier than the previous gen engines of the 737NG/320ceo]
Plus Airbus just has a completely different design approach and architecture for flight control. Airbus A320 was designed as a fly by wire aircraft, from the get-go, the Boeing 737 was designed in a different era, and manual mode had no real computer control (except automatic trims & MCAS). Additionally, there may be differences in the sensor/computer architecture etc.
Bottom line, the 737Max needed a new system, while the A320neo probably needed change in configurations but no new system for control...

This is the correct answer. At least the first 3 paragraphs are, the B737 has Flight Control Computers.

No. The reason the 737 MAX had issues was because of the POSITION of the engines. There's nothing inherently wrong with larger engines on the same airframe. The reason the 737 MAX had issues was because the engines needed to be moved forward and up. The reason for that is that the 737 has short landing gear. Even with CFM-56 engines (737 Classic/NG), there's not much ground clearance between the nacelle and the ground. If you compare a 737 (any model) with an A320-series aircraft, you'll see that the 737 rides a LOT closer to the ground than its Airbus counterpart. You may ask yourself why this is, and to see that, we need to take a look at the history and original mission of the 737. The 737 was originally intended to perform a role similar to what we today know as a regional jet. It was intended to carry relatively small numbers of passengers to smaller regional airports. These smaller airports sometimes lacked the necessary ground service equipment to load an aircraft sitting higher off the ground. In fact, aircraft were sometimes loaded by hand. In order to accommodate this, Boeing equipped the aircraft with shorter landing gear. The original JT8D engines could fit under the airplane easily.
When Boeing decided to equip the 737 with modern engines, they chose the CFM-56. However, the CFM-56 was too big to fit under the wing with adequate ground clearance. Boeing worked with CFM to modify it, and the bottom of the nacelle was flattened. This gave the 737 Classic/NG its unique look. While there was enough ground clearance between the bottom of the engines and the ground, it wasn't much. When Boeing decided to re-engine the re-engined 737, they chose the LEAP-1B. Once again, there was not enough ground clearance with the new engines. Though all this, Boeing had kept the original short landing gear, both for weight savings and the fact that lengthening the gear would require a significant redesign of the airframe structure. There was no other place to put the engines other than moving them forward and up. The LEAP-1B also received a flat bottom. With these modifications, the aircraft had enough ground clearance. Unfortunately, the new engine placement resulted in altered high-alpha flight characteristics. This was enough of a difference that Boeing found it necessary to add a software "fix" that would give the aircraft a slow-flight feel more like that of the 737NG. The A320 came about much later than the 737, and airports by then had the necessary GSE to service taller airplanes. The A320 was then given longer landing gear, and an unmodified CFM-56 fit easily under the wing. The reason that the A320NEO does not need a system like MCAS is because the engines have the necessary clearance under the wing and are in a position that does not negatively affect flight characteristics.